A particle of mass 5 m initially at rest explodes into three fragments with mass ratio g : 1 : 1. Two of the fragments each of mass m are found to move with a speed 60 m/s in mutually perpendicular directions.Then velocity of the third fragment (in m/s) will be
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a
60 2
b
203
c
10 2
d
20 2
answer is D.
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Detailed Solution
I€t the velocity of third fragment b€ v. Applying the law of conservation of momentum, we have(m×60)2+(m×60)2=(3m×v)2 7200m2=9m2v2 V=202m/s